-0.381 966 011 254 48 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 254 48(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.381 966 011 254 48(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.381 966 011 254 48| = 0.381 966 011 254 48


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.381 966 011 254 48.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.381 966 011 254 48 × 2 = 0 + 0.763 932 022 508 96;
  • 2) 0.763 932 022 508 96 × 2 = 1 + 0.527 864 045 017 92;
  • 3) 0.527 864 045 017 92 × 2 = 1 + 0.055 728 090 035 84;
  • 4) 0.055 728 090 035 84 × 2 = 0 + 0.111 456 180 071 68;
  • 5) 0.111 456 180 071 68 × 2 = 0 + 0.222 912 360 143 36;
  • 6) 0.222 912 360 143 36 × 2 = 0 + 0.445 824 720 286 72;
  • 7) 0.445 824 720 286 72 × 2 = 0 + 0.891 649 440 573 44;
  • 8) 0.891 649 440 573 44 × 2 = 1 + 0.783 298 881 146 88;
  • 9) 0.783 298 881 146 88 × 2 = 1 + 0.566 597 762 293 76;
  • 10) 0.566 597 762 293 76 × 2 = 1 + 0.133 195 524 587 52;
  • 11) 0.133 195 524 587 52 × 2 = 0 + 0.266 391 049 175 04;
  • 12) 0.266 391 049 175 04 × 2 = 0 + 0.532 782 098 350 08;
  • 13) 0.532 782 098 350 08 × 2 = 1 + 0.065 564 196 700 16;
  • 14) 0.065 564 196 700 16 × 2 = 0 + 0.131 128 393 400 32;
  • 15) 0.131 128 393 400 32 × 2 = 0 + 0.262 256 786 800 64;
  • 16) 0.262 256 786 800 64 × 2 = 0 + 0.524 513 573 601 28;
  • 17) 0.524 513 573 601 28 × 2 = 1 + 0.049 027 147 202 56;
  • 18) 0.049 027 147 202 56 × 2 = 0 + 0.098 054 294 405 12;
  • 19) 0.098 054 294 405 12 × 2 = 0 + 0.196 108 588 810 24;
  • 20) 0.196 108 588 810 24 × 2 = 0 + 0.392 217 177 620 48;
  • 21) 0.392 217 177 620 48 × 2 = 0 + 0.784 434 355 240 96;
  • 22) 0.784 434 355 240 96 × 2 = 1 + 0.568 868 710 481 92;
  • 23) 0.568 868 710 481 92 × 2 = 1 + 0.137 737 420 963 84;
  • 24) 0.137 737 420 963 84 × 2 = 0 + 0.275 474 841 927 68;
  • 25) 0.275 474 841 927 68 × 2 = 0 + 0.550 949 683 855 36;
  • 26) 0.550 949 683 855 36 × 2 = 1 + 0.101 899 367 710 72;
  • 27) 0.101 899 367 710 72 × 2 = 0 + 0.203 798 735 421 44;
  • 28) 0.203 798 735 421 44 × 2 = 0 + 0.407 597 470 842 88;
  • 29) 0.407 597 470 842 88 × 2 = 0 + 0.815 194 941 685 76;
  • 30) 0.815 194 941 685 76 × 2 = 1 + 0.630 389 883 371 52;
  • 31) 0.630 389 883 371 52 × 2 = 1 + 0.260 779 766 743 04;
  • 32) 0.260 779 766 743 04 × 2 = 0 + 0.521 559 533 486 08;
  • 33) 0.521 559 533 486 08 × 2 = 1 + 0.043 119 066 972 16;
  • 34) 0.043 119 066 972 16 × 2 = 0 + 0.086 238 133 944 32;
  • 35) 0.086 238 133 944 32 × 2 = 0 + 0.172 476 267 888 64;
  • 36) 0.172 476 267 888 64 × 2 = 0 + 0.344 952 535 777 28;
  • 37) 0.344 952 535 777 28 × 2 = 0 + 0.689 905 071 554 56;
  • 38) 0.689 905 071 554 56 × 2 = 1 + 0.379 810 143 109 12;
  • 39) 0.379 810 143 109 12 × 2 = 0 + 0.759 620 286 218 24;
  • 40) 0.759 620 286 218 24 × 2 = 1 + 0.519 240 572 436 48;
  • 41) 0.519 240 572 436 48 × 2 = 1 + 0.038 481 144 872 96;
  • 42) 0.038 481 144 872 96 × 2 = 0 + 0.076 962 289 745 92;
  • 43) 0.076 962 289 745 92 × 2 = 0 + 0.153 924 579 491 84;
  • 44) 0.153 924 579 491 84 × 2 = 0 + 0.307 849 158 983 68;
  • 45) 0.307 849 158 983 68 × 2 = 0 + 0.615 698 317 967 36;
  • 46) 0.615 698 317 967 36 × 2 = 1 + 0.231 396 635 934 72;
  • 47) 0.231 396 635 934 72 × 2 = 0 + 0.462 793 271 869 44;
  • 48) 0.462 793 271 869 44 × 2 = 0 + 0.925 586 543 738 88;
  • 49) 0.925 586 543 738 88 × 2 = 1 + 0.851 173 087 477 76;
  • 50) 0.851 173 087 477 76 × 2 = 1 + 0.702 346 174 955 52;
  • 51) 0.702 346 174 955 52 × 2 = 1 + 0.404 692 349 911 04;
  • 52) 0.404 692 349 911 04 × 2 = 0 + 0.809 384 699 822 08;
  • 53) 0.809 384 699 822 08 × 2 = 1 + 0.618 769 399 644 16;
  • 54) 0.618 769 399 644 16 × 2 = 1 + 0.237 538 799 288 32;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.381 966 011 254 48(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0101 1000 0100 1110 11(2)

6. Positive number before normalization:

0.381 966 011 254 48(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0101 1000 0100 1110 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.381 966 011 254 48(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0101 1000 0100 1110 11(2) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0101 1000 0100 1110 11(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 1010 0001 0110 0001 0011 1011(2) × 2-2


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -2

Mantissa (not normalized):
1.1000 0111 0010 0010 0001 1001 0001 1010 0001 0110 0001 0011 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-2 + 2(11-1) - 1 =

(-2 + 1 023)(10) =

1 021(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1021(10) =

011 1111 1101(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 1000 0111 0010 0010 0001 1001 0001 1010 0001 0110 0001 0011 1011 =

1000 0111 0010 0010 0001 1001 0001 1010 0001 0110 0001 0011 1011


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1101

Mantissa (52 bits) =
1000 0111 0010 0010 0001 1001 0001 1010 0001 0110 0001 0011 1011


Decimal number -0.381 966 011 254 48 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 1010 0001 0110 0001 0011 1011

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100